Prove that the diagonals of a regular octagon intersect at the angular points of a square. $ABCDEFGH$ is a regular octagon and $AF, BE, CH, DG$ are drawn. Prove that their intersections are the angular points of a square.

This question is pretty trivial to solve with coordinate geometry, but it is time-consuming to find the coordinates. Can anybody please provide a solution that does not involve coordinate geometry?
 A: The octogon is invariant under a rotation of angle $\pi/2$ about its center $O$, with diagonals being transformed as $AF \to CH \to EB \to GD \to AF$. Then the quadrilateral with vertices at the intersections of those diagonals is also invariant under a rotation of angle $\pi/2$ around $O$, hence it is a square.
A: Another way.
Let $BE\cap CH=\{K\},$ $BE\cap DG=\{L\}$, $AF\cap DG=\{M\}$ and $AF\cap CH=\{N\}.$
Thus, since our octagon is cyclic, we obtain: $$\measuredangle EBF=\measuredangle BFA$$ and from here  $$BE||AF,$$ which gives $$KL||MN.$$
Similarly $$KN||ML,$$ which gives that $KLMN$ is a parallelogram.
Now, since $$\measuredangle ACH=\measuredangle BAC,$$ we obtain $$AB||CH,$$ which gives $ABKN$ is a parallelogram and from  here $$KN=AB.$$
Similarly, $$KN=CD,$$ which gives that $KLMN$ is a rhombus.
Also, $$\measuredangle KNM=\measuredangle ACH+\measuredangle CAF=\frac{1}{2}\cdot\frac{360^{\circ}}{8}+\frac{1}{2}\cdot3\cdot\frac{360^{\circ}}{8}=90^{\circ},$$ which gives that $KLMN$ is a square.
